evaluate the following using remainder theorem:
f(x)= x³ + 2x² – 3x – 8; f (½)

[tex]f(x)= {x}^{3} + 2 {x}^{2} - 3x - 8 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: f \small{\bigg( \frac{1}{2} \bigg) } = \bigg ( \frac{1}{2} \bigg )^{3} + 2 \bigg( \frac{1}{2} \bigg )^{2} - 3 \bigg( \frac{1}{2} \bigg) - 8 \\ f \small{\bigg( \frac{1}{2} \bigg) } = \frac{1}{8} + 2 \bigg( \frac{1}{4} \bigg) - \frac{3}{2} - 8 \: \: \\f \small{\bigg( \frac{1}{2} \bigg) } = \frac{1}{8} + \frac{2}{4} - \frac{3}{2} - 8 \: \: \: \: \: \: \: \: \: \: \: \\ f \bigg( \frac{1}{2} \bigg ) = \frac{1}{8} + \frac{4}{8} - \frac{12}{8} - \frac{64}{8} \: \: \\ f \bigg( \frac{1}{2} \bigg ) = \frac{5}{8} - \frac{76}{8} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ f \bigg( \frac{1}{2} \bigg) = \frac{ - 71}{ \: \: \: 8} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]

Rheyannpoitogo Rheyannpoitogo · Answer 2

[tex]f \binom{1}{2} = \frac{ - 71}{8} [/tex]

Step-by-step explanation:

[tex]f(x) = x {}^{3} + 2x {}^{2} - 3x - 8[/tex]

[tex]f \binom{1}{2} = \binom{1}{2} = + 2\binom{1}{2} - 3 \binom{1}{2} [/tex]

[tex]f \binom{1}{2} = \frac{1}{8} + 2 \binom{1}{4} - \frac{3}{2} - 8[/tex]

[tex]f \binom{1}{2} = \frac{1}{8} + \frac{2}{4 } - \frac{3}{2} - 8[/tex]

[tex]f \binom{1}{2} = \frac{1}{8} + \frac{4}{8} - \frac{12}{8} - \frac{64}{8} [/tex]

[tex]f \binom{1}{2} = \frac{5}{8} - \frac{76}{8} [/tex]

[tex]f \binom{1}{2} = \frac{ - 71}{8} [/tex]